**Indian Army GD Maths Online Test Paper :**For all those thousands of candidates who dream to join Indian army as soldiers, we are here with some really good collection of Maths question to help them clear the written exam. Most of the candidates face problem in this section and so, our experts have jotted out these important questions from previous year paper and sample paper. This **online test** will absolutely help all of those who are preparing for **Army GD post. **

The **Maths section** is quite important as it comprises 30% of the of the written exam. Since the syllabus is vast it becomes quite necessary to give the online test to practice all types of question. As much as it is important to work hard and practice daily you also need to give this online test in order to score good marks. Giving online test is the best way to practice for any exam.

See also – Indian Army GD Online Test – General Knowledge

## Indian Army GD Maths Online Test Paper

**Q 1.**If 36 men can do a piece of work in 25 hours, in how many hours will 15 men do it?

**60**
- 55
- 62
- 58
- Answer is: 60

**Answer** With Explanation
- Let the required number of hours be x

Then, 15 × x = 36 × 25

x = (36 × 25)/15=60

Hence, 15 men can do it in 60 hours.

**Q 2.**The ratio of third proportional to 12 and 30 and the mean proportional between 9 and 25 is

- 2:1
**5:1**
- 7:15
- 9:14
- Answer is: 5:1

**Answer** With Explanation
- Let, the third proportional to 12 and 3o be x

Then, 12 : 30 :: 30 : x

12x = 30 x 30

x=(30 x 30)/12=75

∴ Third proportional = 75

Mean proportional between 9 and 25 = √9×25 = 15

∴ Required Ratio = 75 : 15 = 5 : 1

**Q 3.**A car is running at a speed of 108 km/h. What distance will it cover in 15 Seconds?

- 45
- 55
**450**
- 455
- Answer is: 450

**Answer** With Explanation
- Speed of car = 108 km/h

=(108 x 1000)/(60 x 60)

= 30 m/s

Time taken = 15 sec

Distance covered = speed x time

= 30 x 15

= 450 m.

**Q 4.**If the radius of a circle is increased by 75% then its circumference will increase by

- 20%
- 50%
**75%**
- 100%
- Answer is: 75%

**Answer** With Explanation
- Let original radius be R cm.

Then, original circumference = (2 πR)cm

New radius = (175% of R) cm

=(175/100 x R)cm=7R/4 cm

New circumference = (2 π x 7R/4)cm

= 7πR/2 cm.

Increase in circumference = (7πR/2-2πR)cm

=3πR/2 cm.

Increase in % = (3πR/2 x 1/2πR x 100)%=75%

**Q 5.**If two angles of a triangle are 30 and 105, then the third angle will be

- 75
**45**
- 65
- 60
- Answer is: 45

**Answer** With Explanation
- Given two angles of triangle

∠1= 30°

∠2= 105°

Let the third angle be x

Then,

Sum of all angles of triangle = 180°

30°+ 105°+x= 180°

x= 180°- 135°

x=45°

∴third angle of ∆= 45°.

**Q 6.**The HCF and LCM of two numbers are 4 and 48 respectively. If one of those numbers is 12, then other number is

- 9
- 8
**12**
- 16
- Answer is: 12

**Answer** With Explanation
- Given, HCF = 4

LCM = 48

One number is 12

HCF×LCM = Product of numbers

Then, other number = (4×48)/12

= 16

**Q 7.**Perimeter of a square is 1000 cm. Find one side of that square.

- 10cm
- 100cm
- 20cm
**250cm**
- Answer is:

**Answer** With Explanation
- Given, Perimeter of square = 1000 cm

Let, side of square = x cm

Then, formula of perimeter of square = 4x = 4×side

∴ 4x = 1000

X = 1000/4

X= 250 cm

**Q 8.**If the average of 40, 10, 25, 20, 35 and x is 25, then the value of x is

**20**
- 30
- 35
- 36
- Answer is: 20

**Answer** With Explanation
- Average= (Sum of all numbers)/(number of terms)

25=(40+10+25+20+35+x)/6

25=(130+6)/6

25×6 = 130+6

150 – 130 = x

X = 20

**Q 9.**. Simplify

81 x 81 + 68 x 68 – 2 x 81 x 68

**169**
- 196
- 201
- None of these
- Answer is: 169

**Answer** With Explanation
- 81×81 + 68×68 – 2×81 ×68

(81)^2+(68)^2-2(81)(68)

Comparing to formula

(a-b)^2=a^2+b^2-2ab

(81-68)^2=(81)^2+(68)^2-2(81)(68)

(81-68)^2=(13)^2

=169

**Q 10.**In 10 years, A will be twice as old as B was 10 years ago, If A is now 9 years older than B, the present age of B is

- 19
- 29
**39**
- 49
- Answer is: 39

**Answer** With Explanation
- Let, B’s present age = x

Then, A’s present age = x+9

10 years ago, B’s age = x – 10

10 years later, A’s age = x+9+10 = 19+x

19+x = 2(x-10)

19+x = 2x – 20

19+20 = 2x-x

39 = x

∴ B’s age = 39 years

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